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Does the series converge? $$\sum_{n=2}^{\infty}\frac{(-1)^n}{\cos{\frac{\pi}{2n}}}$$

I tried to solve it with alternating series test and got $\lim_{n\rightarrow\infty}a_n=1\neq0\implies\text{??Divergence?? or how to prove that}$

Idea: To show that the limit of $(-1)^n$ doesnt exist, thus the series is divergent. Is this correct?

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3 Answers 3

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You're correct, this series is not convergent because sequence $$ \frac{(-1)^n}{\cos\frac{\pi}{2n}}$$ is not convergent. For a series $\sum_n {a_n}$ to be convergent, it is necessary (but not yet sufficient) that $\lim_{n\to\infty} a_n$ exists and is equal to $0$.

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  • $\begingroup$ and because this limit doesn't tend to zero it is sufficient to say that the series diverges? $\endgroup$
    – VLC
    Feb 27, 2021 at 21:08
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    $\begingroup$ @BiliDebili the vanishing condition states that if a series is summable then $lim_{n \to \infty} a_n =0$. Since the series in question doesn't (the consequent, the latter condition) then by modus tollens the series is not summable. $\endgroup$
    – Simone
    Feb 27, 2021 at 21:12
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    $\begingroup$ This sequence doesn't have a limit, in particular it doesn't tend to zero, which is sufficient to tell that the series does not converge. $\endgroup$ Feb 27, 2021 at 21:18
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Series diverges, as general member doesn't tends to zero.

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Think about the behavior of $\cos\frac{\pi}{2n}$ for many $n$'s. Then evaluate the partial sums as $n$ increases.

It's a bit of a trick question.

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